如何用一个两重循环实现行和列的检查

<p>for(i = 0; i &lt; size; i ++){</p><p> numOfO = numOfX = 0;</p><p> //遍历列</p><p> for(k = 0; k &lt; size; k ++){</p><p> if(board[k][i] == 1)</p><p> numOfX ++;</p><p> else</p><p> numOfO ++;</p><p> }</p><p> if(numOfO == size){</p><p> result = 0;</p><p> break;</p><p> }</p><p> else if(numOfX == size){</p><p> result = 1;</p><p> break;</p><p> }</p><p> //遍历行</p><p> for(j = 0; j &lt; size; j ++){</p><p> if(board[i][j] == 1)</p><p> numOfX ++;</p><p> else</p><p> numOfO ++;</p><p> }</p><p> if(numOfO == size){</p><p> result = 0;</p><p> break;</p><p> }</p><p> else if(numOfX == size){</p><p> result = 1;</p><p> break;</p><p> }</p><p>}</p><p><br ></p>

for(i=0;i numOfO = numOfX = 0;//遍历列 for(k = 0; k &lt; size k p&gt; if(board[k][i] == 1) numOfX++; else numOfO ++; if(numOfO == size) [ result = 0; break; else if(numOfX == size)( result= 1; break; 1/遍历行 for(j=0;j if(board[i][]== 1) numOfX++; else numOfO++; if(numOfO == size)[ result =0; break; else if(numOfX == size)(result=1; break; ))

for(i=0;inumOfO = numOfX = 0;//遍历列for(k = 0; k if(board[k][i] == 1)numOfX ;elsenumOfO ;if(numOfO == size) [result = 0;break;else if(numOfX == size)( result= 1;break;1/遍历行for(j=0;jif(board[i][]== 1)numOfX ;elsenumOfO ;if(numOfO == size)[result =0;break;else if(numOfX == size)(result=1;break;))

for(i=0;inumOfO = numOfX = 0;//遍历列for(k = 0; k if(board[k][i] == 1)numOfX ;elsenumOfO ;if(numOfO == size) [result = 0;break;else if(numOfX == size)( result= 1;break;1/遍历行for(j=0;jif(board[i][]== 1)numOfX ;elsenumOfO ;if(numOfO == size)[result =0;break;else if(numOfX == size)(result=1;break;))

for(i=0;inumOfO = numOfX = 0;//遍历列for(k = 0; k if(board[k][i] == 1)numOfX ;elsenumOfO ;if(numOfO == size) [result = 0;break;else if(numOfX == size)( result= 1;break;1/遍历行for(j=0;jif(board[i][]== 1)numOfX ;elsenumOfO ;if(numOfO == size)[result =0;break;else if(numOfX == size)(result=1;break;))

for(i=0;inumOfO = numOfX = 0;//遍历列for(k = 0; k if(board[k][i] == 1)numOfX ;elsenumOfO ;if(numOfO == size) [result = 0;break;else if(numOfX == size)( result= 1;break;1/遍历行for(j=0;jif(board[i][]== 1)numOfX ;elsenumOfO ;if(numOfO == size)[result =0;break;else if(numOfX == size)(result=1;break;))

for(i=0;inumOfO = numOfX = 0;//遍历列for(k = 0; k if(board[k][i] == 1)numOfX ;elsenumOfO ;if(numOfO == size) [result = 0;break;else if(numOfX == size)( result= 1;break;1/遍历行for(j=0;jif(board[i][]== 1)numOfX ;elsenumOfO ;if(numOfO == size)[result =0;break;else if(numOfX == size)(result=1;break;))

for(i=0;inumOfO = numOfX = 0;//遍历列for(k = 0; k if(board[k][i] == 1)numOfX ;elsenumOfO ;if(numOfO == size) [result = 0;break;else if(numOfX == size)( result= 1;break;1/遍历行for(j=0;jif(board[i][]== 1)numOfX ;elsenumOfO ;if(numOfO == size)[result =0;break;else if(numOfX == size)(result=1;break;))

for(i=0;inumofO=numofx=0J/遍历列for(k= 0;k if(board[k[i]==1)numOfX;elsenumofO ;f(numof0 == size) [result= 0breakelse if(numofx== size)( result= 1;break;1/遍历行for(=0;jif(boardill== 1)numOfX;elsenumofo f(numofo == size)[result =0;break;else if(numOfX== size)(result=1;break))

<p>可以使用以下代码实现行和列的检查：</p><p><br ></p><p>int i, j;</p><p>int matrix[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};</p><p><br ></p><p>// 检查行</p><p>for (i = 0; i &lt; 3; i++) {</p><p>&nbsp; &nbsp; int sum = 0;</p><p>&nbsp; &nbsp; for (j = 0; j &lt; 3; j++) {</p><p>&nbsp; &nbsp; &nbsp; &nbsp; sum += matrix[i][j];</p><p>&nbsp; &nbsp; }</p><p>&nbsp; &nbsp; if (sum == 15) {</p><p>&nbsp; &nbsp; &nbsp; &nbsp; printf(&quot;第 %d 行满足条件\n&quot;, i + 1);</p><p>&nbsp; &nbsp; }</p><p>}</p><p><br ></p><p>// 检查列</p><p>for (j = 0; j &lt; 3; j++) {</p><p>&nbsp; &nbsp; int sum = 0;</p><p>&nbsp; &nbsp; for (i = 0; i &lt; 3; i++) {</p><p>&nbsp; &nbsp; &nbsp; &nbsp; sum += matrix[i][j];</p><p>&nbsp; &nbsp; }</p><p>&nbsp; &nbsp; if (sum == 12) {</p><p>&nbsp; &nbsp; &nbsp; &nbsp; printf(&quot;第 %d 列满足条件\n&quot;, j + 1);</p><p>&nbsp; &nbsp; }</p><p>}</p><p><br ></p><p>上述代码中，我们通过两重循环遍历二维数组 `matrix` 中的所有元素，分别计算每行和每列的元素之和。如果行的元素之和等于 15，则输出该行满足条件；如果列的元素之和等于 12，则输出该列满足条件。需要注意的是，这里的检查条件是根据题目要求进行设定的，实际应用中可能会有所不同。</p><p><br ></p>

<p>利用嵌套循环，分别可以利用行-列嵌套遍历每行中的各个元素、利用列-行嵌套遍历每列中的元素。</p>

利用嵌套循环，分别可以利用行-列嵌套遍历每行中的各个元素、利用列-行嵌套遍历每列中的元素。

<p>for(i=0;inumOfO = numOfX = 0;//遍历列for(k = 0; k &nbsp;if(board[k][i] == 1)numOfX &nbsp;;elsenumOfO &nbsp; ;if(numOfO == size) [result = 0;break;else if(numOfX == size)( result= 1;break;1/遍历行for(j=0;jif(board[i][]== 1)numOfX &nbsp;;elsenumOfO &nbsp;;if(numOfO == size)[result =0;break;else if(numOfX == size)(result=1;break;))</p>

利用嵌套循环，分别可以利用行-列嵌套遍历每行中的各个元素、利用列-行嵌套遍历每列中的元素。

利用嵌套循环，分别可以利用行-列嵌套遍历每行中的各个元素、利用列-行嵌套遍历每列中的元素。

利用嵌套循环，分别可以利用行-列嵌套遍历每行中的各个元素、利用列-行嵌套遍历每列中的元素。

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<p>双重for循环，第一重遍厉行，第二重遍历列</p>

用嵌套循环